MaxWong

There is a \(\pm\) button under the '2' button. That is your +/-

\(\dfrac{16-48}{-2}\\ =\dfrac{48-16}{2}\\ =\dfrac{32}{2}\\ =16\)

\(\dfrac{log_3\frac{3}{4}+log_3\frac{4}{5}+log_3\frac{5}{6}+log_3\frac{6}{7}+log_3\frac{7}{8}+log_3\frac{8}{9}}{log\frac{4}{5}+log\frac{5}{6}+log\frac{6}{7 }+log\frac{7}{8}+log\frac{8}{9}+log\frac{9}{10}}\)

=\(\dfrac{log_33-log_34+log_34-log_35+log_35-log_36+log_36-log_37+log_37-log_38+log_38-log_39}{log4-log5+log5-log6+log6-log7+log7-log8+log8-log9+log9-log10}\)= \(\dfrac{log_33-log_39}{log4-log10}\)

=\(\dfrac{1-2}{\log\dfrac{2}{5}}\)

=\(-\log\dfrac{2}{5}\)

Same as CPhill's answer.

\(10^2\div 0.0541\\ =100 \div \dfrac{541}{10000}\\ = 100 \times \dfrac{10000}{541}\\ =\dfrac{1000000}{541}\\ \approx 1828.429\)

http://web2.0calc.com/questions/i-couldn-t-find-my-question-so#r1

I am using this too :D I love this website that I can teach people to do Maths and being taught to do maths :D

It turns into b***s if you type b***s. That actually is B a L L s

If Bowser drew a green one, which means he is lying......

There is 75% that Mario's initial draw is red

And 66 2/3 % that Mario's second draw is red.

Multiply them together = 50%.

There are 80% chance that he has a 50% probability to draw 2 red ones in a row.

Therefore 80% x 50% = 40%

He has a 40% probability now.

If Bowser drew a red one, which means he isn't lying.....

There is 50% that Mario's initial draw is red

And 33 1/3 % that Mario's second draw is red.

Multiply them together = 16 1/6 %

There are 20% chance that he has a 16 1/6% probability to draw 2 red ones in a row.

Therefore 20% x 16 1/6% = 3 1/3%

He has a 40% + 3 1/3% probability now.

Therefore Mario has a 43 1/3% probability to draw 2 red b***s in a row.

On calculator: a/b 

In LaTeX: \dfrac{a}{b} and \frac{a}{b} where the \dfrac command makes bigger fractions.

e.g. : \(\dfrac2{3}\) Using \dfrac

\(\frac{2}{3}\) Using \frac

\(\dfrac{3}{8}(16x-48)+3x\\ =3(2x-6)+3x\\ = 6x - 18 + 3x\\ = 9x - 18\)

If {a,b} and {b,a} is counted the same.......

Starting with 1:

{1,3}, {1,4}, ...... {1,9}, {1,3,5}, {1,3,6}...... ,{1,3,9}......

Subtotal 7+5+3+1 subsets = 16 subsets.

Starting with 2:

{2,4}, .......{2,9}, {2,4,6}, {2,4,7}... {2,4,9}...... Too lazy to type them all XD

Subtotal 6 + 4 + 2 = 12 subsets

Starting with 3: 

{3,5}, {3,6} .... {3,9}, {3,5,7}, ..., {3 , 5, 9}.........

Subtotal 5 + 3 + 1 = 9 subsets

There we have found a rule for this.

Total subsets = 4 x 4 + 4 x 3 + 3 x 3 + 3 x 2 + 2 x 2 + 2 x 1 + 1 x 1 = 16 + 12 + 9 + 6 + 4 + 2 + 1 = 50 subsets.